Simplify matrix algebra equation. I am reading Kuipers book's of Rotation sequences, in chapter 11 he develop a method to calculate the perturbations in a euler rotation sequence, in this case the $ZYX$ sequence
He defines a pertubation rotation sequence, $R =(\Delta_\psi,\Delta_\theta,\Delta_\phi)$
And the rotation frame sequence $T = (\psi,\theta,\phi)$
more specifically
$T = T_\phi*T_\theta*T_\psi$ is the euler rotation sequence $(\psi,\theta,\phi)$ that rotates the $XYZ$ frame in the order: first $Z$ by $\psi$, $Y'$ by $\theta$ and $X''$ by $\phi$
the same reasoning goes to $R$
So $S = RT$ is the full sequence that rotates the $XYZ$ frame in a perturbed rotated frame. Knowing this we can develop a closed rotate sequence that gives us the identity matrix
$I = RTT^TR^T = R^TT^TRT = S^TRT=RTS^T$
Now he introduces this
$R T_\phi T_\theta T_\psi = S_\psi S_\theta S_\phi $
where $S$ can be also viewed as $T$ but where each angle has now added the deltas values (perturbations values) (also I don't understand why he reverse the order of the angles)
$R T_\phi T_\theta T_\psi = T_{\phi+\Delta_\phi} T_{\theta+\Delta_\theta} T_{\psi+\Delta_\psi}  $
This is also the same as
$R T_\phi T_\theta T_\psi = T_{\phi} T_{\Delta_\phi} T_{\theta} T_{\Delta_\theta}  T_{\psi} T_{\Delta_\psi}  $
Now he start the matrix algebra process
$$\eqalign{
R T_\phi T_\theta T_\psi &= T_{\phi} T_{\Delta_\phi} T_{\theta} T_{\Delta_\theta}  T_{\psi} T_{\Delta_\psi} \\
R T_\phi T_\theta T_\psi(T_{-\psi}) &= T_{\phi} T_{\Delta_\phi} T_{\theta} T_{\Delta_\theta}  T_{\psi} T_{\Delta_\psi}(T_{-\psi}) \\
R T_\phi T_\theta &= T_{\phi} T_{\Delta_\phi} T_{\theta} T_{\Delta_\theta}  T_{\Delta_\psi}
}$$
Now he introduces matrix $A = T_\phi T_\theta$
$$\eqalign{
R A &= T_{\phi} T_{\Delta_\phi} T_{\theta} T_{\Delta_\theta}  T_{\Delta_\psi} \\
R A &= (T_{\phi} T_{\Delta_\phi}) T_{\theta} T_{\Delta_\theta}  T_{\Delta_\psi} \\
R A &= (T_{\Delta_\phi} T_{\phi}) T_{\theta} T_{\Delta_\theta}  T_{\Delta_\psi} \\
R A &= T_{\Delta_\phi} (T_{\phi} T_{\theta}) T_{\Delta_\theta}  T_{\Delta_\psi} \\
R A &= T_{\Delta_\phi} A T_{\Delta_\theta}  T_{\Delta_\psi} \\
R A A^T &= T_{\Delta_\phi} A T_{\Delta_\theta}  T_{\Delta_\psi} A^T \\
R &= T_{\Delta_\phi} A T_{\Delta_\theta}  T_{\Delta_\psi} A^T \\
}$$
Now here is the complicated stuff
$T_{\Delta_\phi} = I + \Phi$ where $\Phi = \begin{vmatrix}
0 & 0 & 0 \\ 
0 & 0 & \Delta\phi \\
0 & -\Delta\phi & 0
\end{vmatrix}$
$T_{\Delta_\theta} = I + \Theta$ where $\Theta = \begin{vmatrix}
0 & 0 & -\Delta\theta \\ 
0 & 0 & 0 \\
\Delta\theta & 0 & 0
\end{vmatrix}$
$T_{\Delta_\psi} = I + \Psi$ where $\Psi = \begin{vmatrix}
0 & \Delta\psi & 0 \\ 
-\Delta\psi & 0 & 0 \\
0 & 0 & 0
\end{vmatrix}$
$R = I + \Omega$ where $\Omega = \begin{vmatrix}
0 & \Delta_w & -\Delta_v \\ 
-\Delta_w & 0 & \Delta_u \\
\Delta_v & -\Delta_u & 0
\end{vmatrix}$
$\Delta_u = $ roll increment $\Delta_x$
$\Delta_v = $ pitch increment $\Delta_y$
$\Delta_w = $ yaw increment $\Delta_z$
So
$$\eqalign{
R &= T_{\Delta_\phi} A T_{\Delta_\theta}  T_{\Delta_\psi} A^T  \\
I + \Omega &= [I + \Phi] A [I + \Theta][I + \Psi] A^T \\
I + \Omega &= [I + \Phi] A [I + \Theta + \Psi] A^T \\
I + \Omega &= [I + \Phi][I + A \Theta A^T + A \Psi A^T]
}$$
NOW HERE IS MY PROBLEM, in the book he concludes that this equation ends like this
$$\eqalign{
\Omega &= \Phi + A [ \Theta + \Psi] A^T
}$$
But no matter what my approach is, I can't simplify that equation to obtain that result!
If a distribute the $[I + \Phi]$ term I'll obtain this
$$\eqalign{
I + \Omega &= [I + \Phi][I + A \Theta A^T + A \Psi A^T] \\
&= [I + A \Theta A^T + A \Psi A^T] + [\Phi + \Phi A \Theta A^T + \Phi A \Psi A^T] \\
&= (I + A [ \Theta + \Psi] A^T) + (\Phi + \Phi A [ \Theta + \Psi] A^T) \\
\Omega &= A [ \Theta + \Psi] A^T + \Phi + \Phi A [ \Theta + \Psi] A^T \\
&= \Phi + A [ \Theta + \Psi] A^T + \Phi A [ \Theta + \Psi] A^T \\
&= \Phi + [I + \Phi](A [ \Theta + \Psi] A^T) \\
&= \Phi + T_{\Delta_\phi} (A [ \Theta + \Psi] A^T) \\
&= \Phi + T_{\Delta_\phi} A \Theta A^T + T_{\Delta_\phi} A \Psi A^T
}$$
I am stuck at this, I cannot going further.
Here is a screenshot from the pdf: https://imgur.com/HbmEVpz
// UPDATE: 16/03/2020
lets start with this monster
$$\eqalign{
&= \Phi + T_{\Delta_\phi} A \Theta A^T + T_{\Delta_\phi} A \Psi A^T
}$$
I assume that WE CAN CONMUTE incremental rotational matrices, because the change i so small
also multiplying incremental rotational = $\Phi\Theta = \Phi\Psi = \Theta\Psi = 0$
also, remember
Δ=+Φ
Δ=+Θ
Δ=+Ψ
lets start with the term $T_{\Delta_\phi} A \Theta A^T$
$$\eqalign{
&= T_{\Delta_\phi} A \Theta A^T \\
&= T_{\Delta_\phi} T_{\phi} T_{\theta} [T_{\Delta_\theta} - I] T_{-\theta} T_{-\phi} \\
&= T_{\Delta_\phi} T_{\phi} (T_{\theta} T_{\Delta_\theta} T_{-\theta})  T_{-\phi} + T_{\Delta_\phi} T_{\phi} T_{\theta} (-I) T_{-\theta} T_{-\phi} \\
&= T_{\Delta_\phi} T_{\phi} T_{\Delta_\theta} T_{-\phi} - T_{\Delta_\phi} A A^T \\
&= T_{\phi} (T_{\Delta_\phi} T_{\Delta_\theta}) T_{-\phi} - T_{\Delta_\phi} \\
&= T_{\phi} T_{\Delta_\theta} (T_{\Delta_\phi} T_{-\phi}) - T_{\Delta_\phi} \\
&= (T_{\phi} T_{\Delta_\theta}) T_{-\phi} T_{\Delta_\phi} - T_{\Delta_\phi} \\
&= T_{\Delta_\theta} (T_{\phi} T_{-\phi} T_{\Delta_\phi}) - T_{\Delta_\phi} \\
&= T_{\Delta_\theta} T_{\Delta_\phi} - T_{\Delta_\phi} \\
&= [\Theta + I][\Phi + I] - [\Phi + I] \\
&= [\Theta \Phi + \Theta + \Phi + I] - [\Phi + I] \\
&= [\Theta + \Phi + I] - [\Phi + I] \\
&= \Theta + \Phi + I - \Phi - I \\
&= \Theta
}$$
continue with $ T_{\Delta_\phi} A \Psi A^T $
$$\eqalign{
&= T_{\Delta_\phi} A \Psi A^T \\
&= [\Phi + I] A \Psi A^T \\
&= \Phi A \Psi A^T + A \Psi A^T \\
&= \Phi A [ T_{\Delta_\psi} - I ] A^T + A [ T_{\Delta_\psi} - I ] A^T \\
&= \Phi A T_{\Delta_\psi} A^T + \Phi A (-I) A^T + A T_{\Delta_\psi} A^T + A (-I) A^T \\
&= \Phi A T_{\Delta_\psi} A^T - \Phi + A T_{\Delta_\psi} A^T - I \\
&= \Phi A [ \Psi + I ] A^T - \Phi + A [ \Psi + I ] A^T - I \\
&= \Phi A \Psi A^T + \Phi A I A^T - \Phi + A \Psi A^T + A A^T - I \\
&= \Phi A \Psi A^T + \Phi - \Phi + A \Psi A^T + I - I \\
&= \Phi A \Psi A^T + A \Psi A^T \\
&= [\Phi + I] A \Psi A^T \\
}$$
as you can see there is no way to simplify this term :(((((
So we got this, we can add zeroes to make prettier, but still no luck!!!
$$\eqalign{
my &\neq book \\
\Phi + \Theta + [\Phi + I] A \Psi A^T &\neq \Phi + A [\Theta + \Psi ] A^T \\
\Phi + \Theta + [\Phi + I] A \Psi A^T + I - I &\stackrel{?}{=} \\
[\Phi + I] + [\Theta - I] + [\Phi + I] A \Psi A^T &\stackrel{?}{=} \\
[\Phi + I][I + A \Psi A^T ] + [\Theta - I] + I - I &\stackrel{?}{=} \\
[\Phi + I][I + A \Psi A^T ] + [\Theta + I] &\stackrel{?}{=} \\
}$$
But still not is the same as the book!! I cant sleep if I dont solved this crazyness
 A: Update 18/03/2020
Well I think that the author discarded the last term because its components when you put some values they give you very small values, the thing is that without Mathematica I could never end to this conclusion. Starting when the author start the algebra process
$$\eqalign{
I+Ω &= [I+Φ][I+AΘA^T+AΨA^T] \\
I+Ω &= [I+AΘA^T+AΨA^T] + Φ[I+AΘA^T+AΨA^T] \\
I+Ω &= [I+A[Θ+Ψ]A^T] + [Φ + ΦA[Θ+Ψ]A^T] \\
I+Ω &= I+A[Θ+Ψ]A^T +  Φ + ΦA[Θ+Ψ]A^T \\
Ω &= A[Θ+Ψ]A^T + Φ + ΦA[Θ+Ψ]A^T \\
}$$
With the help of Mathematica you can compute easly the matrix resulting of the term $ΦA[Θ+Ψ]A^T$ which is 
$ΦA[Θ+Ψ]A^T = \begin{vmatrix}
0 & 0 & 0 \\ 
\Delta_\phi(cos(\phi)\Delta_\theta + cos(\theta)sin(\phi)\Delta_\psi) & sin(\theta)\Delta_\phi\Delta_\psi & 0 \\
-\Delta_\phi(sin(\phi)\Delta_\theta - cos(\theta)cos(\phi)\Delta_\psi) & 0 & sin(\theta)\Delta_\phi\Delta_\psi
\end{vmatrix}$
because we are talking about small incremental angles, like $\Delta = 0.01$ these equations gives very small values, so I think the author just ignore this term and then he ends as this:
$$\eqalign{
Ω &= Φ +  A[Θ+Ψ]A^T \\
}$$
Just for the record the term $A[Θ+Ψ]A^T$ computes to
$A[Θ+Ψ]A^T = \begin{vmatrix}
0 & -sin(\phi)\Delta_\theta + cos(\theta)cos(\phi)\Delta_\psi & -cos(\phi)\Delta_\theta - cos(\theta)sin(\phi)\Delta_\psi \\ 
sin(\phi)\Delta_\theta - cos(\theta)cos(\phi)\Delta_\psi & 0 & -sin(\theta)\Delta_\psi \\
cos(\phi)\Delta_\theta + cos(\theta)sin(\phi)\Delta_\psi & sin(\theta)\Delta_\psi & 0
\end{vmatrix}$
The final matrix $\Omega$ is
$\Omega = \begin{vmatrix}
0 & -sin(\phi)\Delta_\theta + cos(\theta)cos(\phi)\Delta_\psi & -cos(\phi)\Delta_\theta - cos(\theta)sin(\phi)\Delta_\psi \\ 
sin(\phi)\Delta_\theta - cos(\theta)cos(\phi)\Delta_\psi & 0 & \Delta_\phi -sin(\theta)\Delta_\psi \\
cos(\phi)\Delta_\theta + cos(\theta)sin(\phi)\Delta_\psi & -\Delta_\phi +sin(\theta)\Delta_\psi & 0
\end{vmatrix}$
